Question

It is found that on average, 9% of the population recycles its garbage. Under this assumption, if a random sample of 120 households in Iowa City is taken, would you be surprised to find that fewer than 3% of households in the sample recycle their garbage? Use the fact that p-hat has an approximately normal distribution. Group of answer choices I would not be surprised because there are lots of lazy students in Iowa City. I would not be surprised because 3% is not that far away from 9%. I would be surprised because Iowa City is a very clean community which means lots of people recycle their garbage. I would be surprised because if the 9% average is true, the chance of a sample proportion being less than 3% is very small (only 1%). Who cares? It's only garbage!

Answer 1

Complete Question

It is found that on average, 9% of the population recycles its garbage. Under this assumption, if a random sample of 120 households in Iowa City is taken, would you be surprised to find that fewer than 3% of households in the sample recycle their garbage? Use the fact that p-hat has an approximately normal distribution.

Group of answer choices

A I would not be surprised because there are lots of lazy students in Iowa City.

B I would not be surprised because 3% is not that far away from 9%.

C I would be surprised because Iowa City is a very clean community which means lots of people recycle their garbage.

D I would be surprised because if the 9% average is true, the chance of a sample proportion being less than 3% is very small (only 1%). Who cares? It's only garbage!

Answer:

The correct option is D

Step-by-step explanation:

From the question we are told that

    The sample size is  n  = 120

    The  mean of the proportion is  p =  0.09

   

Generally the standard deviation is mathematically represented as

      $\sigma = \sqrt{\frac{p(1-p)}{n} }$

=>   $\sigma = \sqrt{\frac{0.09(1-0.09)}{120 } }$

=>   $\sigma = 0.02612$

Generally the chance that fewer than 3%   of households in the sample recycle their garbage is mathematically represented as

    $P(X < 0.03) = P( \frac{X - p}{\sigma} < \frac{0.03 -0.09}{0.02612} )$

    $P(X < 0.03) = P( Z < -2.297 )$

From the z-table  

     $P( Z < -2.297 ) = 0.01$

=> [tex]P(X < 0.03) = 0.01/tex]

=>   [tex]P(X < 0.03) = 1\% /tex]